i need help understanding how to write half reactions and full reactions?

Question

IN every problem copper is my cathode but the anode changes such as one is copper then iron and lead, magnesium, and so forth. It asks me to complete anode half- reaction and cathode half - reaction and then the cell reaction. I dont really know how to go abou7t this maybe if someone could walk me through one of the above examples such as copper copper or copper iron then i could do the rest. Copper is the cathode in every problem the other metal is the anode? Can someone please explain this to me? Not only do i Have this assignment to turn in soon but it will be on my next test. THank you

Answer 1

In order to figure out who the cathode and who the anode is, you need standard reduction potential values (usually found in the back of your book)

ALLLLLLL of the reactions are written as reductions:

Cu+2 + 2e- -----> Cu (s) Ered = +0.34 V
Zn+2 + 2e- -----> Zn (s) E red = -0.76 V

In order to figure out who will be reduced (the cathode - think RED CAT for reduction at cathode), the species with the larger/more positive E reduction value will STAY the reduction reaction. THe other reaction must therefore be reversed to indicate oxidation and the E value changes sign

Cu+2 + 2e- -----> Cu (s) Ered = +0.34 V
Zn (s) -----> Zn+2 + 2e- E ox = +0.76 V

Ecell = Cu+2 + Zn(s) ----> Cu(s) + Zn+2
Ecell numerically = Ered + Eox = 0.34 V + 0.76 V = 1.10 V

Let's look at another

Cu+2 + 2e- -----> Cu (s) Ered = +0.34 V
Pb+2 + 2e- -----> Pb (s) E red = -0.13 V

Again, Cu has the more positive/larger E reduction value so it stays AS IS
Reverse the other reaction, change the Ered value's sign

Cu+2 + 2e- -----> Cu (s) Ered = +0.34 V
Pb (s) -----> Pb+2 + 2e- E ox = +0.13 V

Add them together Cu+2 + Pb(s) ---> Pb+2 + Cu(s)

Ecell = 0.47 V

Let's look at another

Cu+2 + 2e- -----> Cu (s) Ered = +0.34 V
K+1 + 1e- -----> K (s) E red = -2.93 V

Here we have a slightly different situation: the numbers of electrons are NOT the same. In any balanced redox reaction, the numbers of electrons MUST be the same

So - let's keep the copper reaction as is (it has the larger/more positive E reduction value.) Let's reverse the potassium reaction and change the sign of its E value

Cu+2 + 2e- -----> Cu (s) Ered = +0.34 V
K (s) -----> K+1 + 1e- E ox = +2.93 V

Now, let's make those electrons equal. Multiply the potassium reaction by 2 - BUT DO NOT CHANGE THE NUMERICAL VALUE OF ITS Ered or Eox POTENTIAL!! THE E VALUES ARE INDEPENDENT OF AMOUNT OF SUBSTANCE - SO THEY DO NOT NOT NOT NOT CHANGE!


Cu+2 + 2e- -----> Cu (s) Ered = +0.34 V
2 K (s) -----> 2 K+1 + 2e- E ox = +2.93 V

Cu+2 + 2K (s) ------> 2K+1 + Cu(s)

Ecell = 3.27 V

Hope that helps!

Answer 2

Try to separate them out.